Question

A two digitised number is such that the product of its digit is 15 .if 18 is added to the number the digit interchange their places.find the numbers

Answer 1

Let the ten's digit be x and one's digit be y. 

The number will be 10x + y.

Given product of its digits is 15 

xy = 15

y = 15/x  --- (1).

Given that when 18 is added to the number the digits get interchanged.

10x + y + 18 = 10y + x

9x - 9y + 18 = 0

x - y + 2 = 0   ------- (2) 

Substitute (1) in (2), we get

x - (15/x) + 2 = 0

x^2 + 2x - 15 = 0

x^2 + 5x - 3x - 15 = 0

x(x + 5) - 3(x + 5) = 0

(x+5)(x-3) = 0

x = -5,3.

Since the digits cannot be negative, x = 3.

Substitute x = 3 in (1)

y = 15/3

   = 5.

The number = 10x + y

                     = 10 * 3 + 5

                     = 30 + 5

                     = 35

Hope this help

Answer 2

Let the ten's digit be x and one's digit be y. 

The number will be 10x + y.

Given product of its digits is 15 

xy = 15

y = 15/x  --- (1).

Given that when 18 is added to the number the digits get interchanged.

10x + y + 18 = 10y + x

9x - 9y + 18 = 0

x - y + 2 = 0   ------- (2) 

Substitute (1) in (2), we get

x - (15/x) + 2 = 0

x^2 + 2x - 15 = 0

x^2 + 5x - 3x - 15 = 0

x(x + 5) - 3(x + 5) = 0

(x+5)(x-3) = 0

x = -5,3.

Since the digits cannot be negative, x = 3.

Substitute x = 3 in (1)

y = 15/3

  = 5.

The number = 10x + y

                    = 10 * 3 + 5

                    = 30 + 5

                    = 35